anonymous
  • anonymous
If a , b, and c are in Arithmetic Progression, then the straight line ax + by + c = 0 will always pass through the point: a) (- 1, -2) b) (1, -2) c) (-1 , 2) d) (1, 2)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Please help.
amistre64
  • amistre64
we should know that the slope of the line is: -a/b
amistre64
  • amistre64
0 = -a/b x - c is then what we have to conform to i believe

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anonymous
  • anonymous
I think so.
amistre64
  • amistre64
well, -c/b on the end i spose would be more accurate
amistre64
  • amistre64
when x=0, -c/b = 0 means that c=0 so: y = -a/b x seems like a fair assumption
amistre64
  • amistre64
i gotta re think that :)
amistre64
  • amistre64
ax +by + c = 0 ax + by = -c y = (-ax -c) /b no zero involved .....
asnaseer
  • asnaseer
If a, b and c are in AP, then doesn't this imply: b = a + d c = a + 2d where d is the difference between each term of the AP?
amistre64
  • amistre64
good, good
anonymous
  • anonymous
Then, how to proceed next?
amistre64
  • amistre64
if my line equation is useful; maybe sub in so that it all speaks in a?
asnaseer
  • asnaseer
Then you can rewrite your equation as: ax + (a+d)y + a + 2d = 0 and see for which point this equation holds true?
amistre64
  • amistre64
\[y = \frac{-ax -(a+2d)}{a+d}\] would be the same set up i believe
asnaseer
  • asnaseer
yes it would
anonymous
  • anonymous
So the exact option?
asnaseer
  • asnaseer
Aadarsh: just put each pair of values into the equations and see which one works
amistre64
  • amistre64
a) (- 1, -2) b) (1, -2) c) (-1 , 2) d) (1, 2) trial and error .... \[-2 \ =^? \frac{a -(a+2d)}{a+d}\] \[-2 \ =^? \frac{-a -(a+2d)}{a+d}\] \[2 \ =^? \frac{a -(a+2d)}{a+d}\] \[2 \ =^? \frac{-a -(a+2d)}{a+d}\]
amistre64
  • amistre64
\[2 \ =^? \frac{-a -(a+2d)}{a+d}\] \[2 \ =^? \frac{-a -a-2d}{a+d}\] \[2 \ =^? \frac{-2a-2d}{a+d}\] \[2 \ =^? -2\frac{a+d}{a+d};F\]
amistre64
  • amistre64
-1,-2 would then seem to be appropriate to me
anonymous
  • anonymous
Is it? I wrote (-1, -2), just guessing.
asnaseer
  • asnaseer
amistre: I get a different result. Aadarsh: what do you get?
amistre64
  • amistre64
i got a typo in my cerbral cortex; :) 1,-2 might be better; would have to test it out
asnaseer
  • asnaseer
:) - I concur
anonymous
  • anonymous
Yes, (1, -2) is the only correct answer.

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